Answer to Question #127724 in Statistics and Probability for ameek

Question

A statistical analysis of 1000 long-distance telephone calls made from the headquarters of the ABC company indicates that the length of these calls is normally distributed with population mean is 240 seconds and standard deviation is 40 seconds.

a) What is the probability that a call lasted less than 180 seconds?

Answer 1

The cumulative distribution function of normal distribution is

$F(x) = \frac{1}{2}\left(1+\mathop{\mathrm{erf}}\left(\frac{x-\mu}{\sigma\sqrt2}\right)\right)$

Here $\mu=240$ and $\sigma = 40$, so

$F(x) = \frac{1}{2}\left(1+\mathop{\mathrm{erf}}\left(\frac{x-240}{40\sqrt2}\right)\right)$

The call can't have a negative length, so the bounds of interval should be $(0,180)$. But math says that the bounds of interval should be $(-\infty,180)$. We can use any of them because the difference is negligible, less than $10^{-9}$.

$P = F(180)-F(0) = \frac{1}{2}\left(\mathop{\mathrm{erf}}\left(\frac{-60}{40\sqrt2}\right)-\mathop{\mathrm{erf}}\left(\frac{-240}{40\sqrt2}\right)\right) \approx \frac{1}{2}(-0.8664+1) = 0.0668$